A radioactive isotope decays at such a rate that after 96 min, only `1//8th` of the original amount remains.
The value of `t_(1//2)` of this nuclide is
a. 12 min b. 32 min c. 24 min d. 48 min

Options

A. a. 12 min
B. b. 32 min
C. c. 24 min
D. d. 48 min

Answer

b. 32 min

Explanation

`K = (2.303)/(t) log (a)/(a - x)`
or `(0.693)/(t_(1//2)) = (2.303)/(96) log (a)/((a)/(8))`
Second method
`(t_(1//2))/(t_(1//8)) = (0.3)/(log (a)/(a//8))`
`:. (t_(1//2))/(96) = (0.3)/(log 8)`
`(t//2)/(96) = (0.3)/(log 2^(3))`
`:. t_(1//2) = (0.3 xx 96)/(3 xx log 2) = 32` min